Solve {ccc}{4}&{3}&{-1}{0}&{5}&{2}{10}&{-7}&{8}

$$\left( \begin{array} { c c c } { 4 } & { 3 } & { - 1 } \\ { 0 } & { 5 } & { 2 } \\ { 10 } & { - 7 } & { 8 } \end{array} \right)$$

Calculate Determinant

$$326$$

Steps Using Sarrus' Rule

Find the determinant of the matrix using the method of diagonals.

$$det(\left(\begin{matrix}4&3&-1\\0&5&2\\10&-7&8\end{matrix}\right))$$

Extend the original matrix by repeating the first two columns as the fourth and fifth columns.

$$\left(\begin{matrix}4&3&-1&4&3\\0&5&2&0&5\\10&-7&8&10&-7\end{matrix}\right)$$

Starting at the upper left entry, multiply down along the diagonals, and add the resulting products.

$$4\times 5\times 8+3\times 2\times 10=220$$

Starting at the lower left entry, multiply up along the diagonals, and add the resulting products.

$$10\times 5\left(-1\right)-7\times 2\times 4=-106$$

Subtract the sum of the upward diagonal products from the sum of the downward diagonal products.

$$220-\left(-106\right)$$

Subtract -106 from 220.

$$326$$

Steps Using Expansion by Minors

Find the determinant of the matrix using the method of expansion by minors (also known as expansion by cofactors).

$$det(\left(\begin{matrix}4&3&-1\\0&5&2\\10&-7&8\end{matrix}\right))$$

To expand by minors, multiply each element of the first row by its minor, which is the determinant of the 2\times 2 matrix created by deleting the row and column containing that element, then multiply by the element's position sign.

$$4det(\left(\begin{matrix}5&2\\-7&8\end{matrix}\right))-3det(\left(\begin{matrix}0&2\\10&8\end{matrix}\right))-det(\left(\begin{matrix}0&5\\10&-7\end{matrix}\right))$$

For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the determinant is ad-bc.

$$4\left(5\times 8-\left(-7\times 2\right)\right)-3\left(-10\times 2\right)-\left(-10\times 5\right)$$

Simplify.

$$4\times 54-3\left(-20\right)-\left(-50\right)$$

Add the terms to obtain the final result.

$$326$$

Evaluate

$$\left(\begin{matrix}4&3&-1\\0&5&2\\10&-7&8\end{matrix}\right)$$